--- title: "Settlement with preconsolidation 125 kPa" subject: "Soil Mechanics" chapter: 9 key: "semester_4/soil_mechanics:primary-consolidation-settlement-preconsolidated" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/primary-consolidation-settlement-preconsolidated site: BCT Engineering --- # Settlement with preconsolidation 125 kPa Soil Mechanics · Chapter 9: Compressibility of Soil - Asked in: 2075 Baisakh, 4 marks - Video: [Consolidation_Primary Consolidation Settlement](https://www.youtube.com/watch?v=vp_p_jLBGTM) — Dr. Maria Cecilia Marcos - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/primary-consolidation-settlement-preconsolidated - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-9 ## Question A soil profile is shown in the figure below. If a uniformly distributed load of 50 kPa is applied on the ground surface, having preconsolidation pressure, compression index and recompression index of 125 kPa, 0.36 and 0.06, respectively, calculate the amount of settlement of the clay layer due to primary consolidation. Take $\gamma_w$ = 10 kN/m³. [Figure: Δσ = 50 kN/m² on the surface; sand layer with γ = 16 kN/m³ from 0 to 2 m (water table at 2 m); sand layer $\gamma_{sat}$ = 18 kN/m³ from 2 to 8 m; clay layer $\gamma_{sat}$ = 20 kN/m³ from 8 to 14 m; sand layer below 14 m.] ## Short answer Initial effective stress at clay mid-depth is 110 kPa, preconsolidation pressure 125 kPa, final stress 160 kPa. Using both recompression and compression indices, primary consolidation settlement is about 148 mm. ## Answer **Given:** clay 8 m to 14 m ($H = 6$ m), mid-depth at 11 m; $\sigma_c' = 125$ kPa, $C_c = 0.36$, $C_r = 0.06$, $\Delta\sigma = 50$ kPa, $\gamma_w = 10$ kN/m³. **Assumption:** the initial void ratio of the clay is not given. Taking $G_s = 2.7$ and $\gamma_{sat} = 20$ kN/m³, $\frac{(2.7+e_0)(10)}{1+e_0}=20$ gives $e_0 = 0.70$. ```text 0 - 2 m sand gamma = 16 2 - 8 m sand gsat = 18 (WT at 2 m) 8 - 14 m clay gsat = 20 <- mid at 11 m ``` **Step 1: Initial effective stress at 11 m** $$ \sigma_0' = 16(2) + (18-10)(6) + (20-10)(3) = 32 + 48 + 30 = 110\ \text{kPa} $$ **Step 2: State of the clay.** $\sigma_0' = 110 < \sigma_c' = 125$, so the clay is overconsolidated. Final stress $\sigma_0'+\Delta\sigma = 160 > \sigma_c'$, so the stress path crosses the preconsolidation pressure and both $C_r$ and $C_c$ are needed. **Step 3: Settlement** $$ \begin{aligned} S_c &= \frac{H}{1+e_0}\left[C_r\log_{10}\frac{\sigma_c'}{\sigma_0'} + C_c\log_{10}\frac{\sigma_0'+\Delta\sigma}{\sigma_c'}\right] \\ &= \frac{6000}{1.70}\left[0.06\log_{10}\frac{125}{110} + 0.36\log_{10}\frac{160}{125}\right] \\ &= 3529\,[0.00330 + 0.03859] = 148\ \text{mm} \end{aligned} $$ **Answer: primary consolidation settlement $\approx 148$ mm** (for $e_0 = 0.70$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. sigma0' = 110 kPa and preconsolidation pressure = 125 kPa. What is the clay's state? A. Overconsolidated, because sigma0' < sigma_c' B. Normally consolidated, because sigma0' is near sigma_c' C. Underconsolidated, because sigma0' < sigma_c' D. Normally consolidated, because sigma0' < sigma_c' Answer: A. A clay whose present effective stress is less than its past maximum is overconsolidated. (Key point: Soil state) 2. Why are both C_r and C_c needed? A. Because the clay is normally consolidated B. Final stress 160 kPa exceeds sigma_c', so the stress path crosses the preconsolidation pressure C. Because the load is below sigma_c' D. Because the water table is at 2 m Answer: B. The path follows the recompression line up to 125 kPa then the virgin compression line beyond it. (Key point: Why both indices) 3. Which expression gives sigma0' at 11 m depth? A. 16(2) + 18(6) + 20(3) = 200 kPa B. 16(2) + 8(6) + 10(6) = 140 kPa C. 16(2) + 8(6) + 10(3) = 110 kPa D. 18(2) + 10(6) + 10(3) = 126 kPa Answer: C. Use gamma above the water table and submerged weights (gamma_sat - 10) below, to the mid-depth of the clay. (Key point: Initial effective stress) 4. Which expression is the settlement? A. H/(1+e0) [C_c log(125/110) + C_r log(160/125)] B. H/(1+e0) C_c log(160/110) C. H/(1+e0) C_r log(160/110) D. H/(1+e0) [C_r log(125/110) + C_c log(160/125)] Answer: D. Recompression C_r applies from 110 to 125 and compression C_c from 125 to 160. 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